Boron has only two naturally occurring isotopes. The mass of boron-10 is 10.01294 amu and the mass of boron-11 is 11.00931 amu.
Use the atomic mass of boron to calculate the relative abundance of boron-10.
Atomic mass of boron is 10.81
How do I solve this?
10.81 = (10.01294x) + (11.00931y)
y=1-x
10.81 = (10.01294x) + (11.00931(1-x))
10.81 = 10.01294x + 11.00931 + -11.00931x
10.81 = 10.01294x + 11.00931 + -11.00931x
10.81 = -0.99637x + 11.00931
10.81-11.00931 = -0.99637x + 11.00931-11.00931
-0.2831 = -0.99637x
-0.2831/-0.99637 = 0.99637/-0.99637x
0.28413139697 = x
Boron 10 = 28.41%
Boron 11 = 71.59%(subtract 100%- 28.41%)
To calculate the relative abundance of boron-10, we can use the formula:
Relative abundance of boron-10 = (mass of boron-10 / atomic mass of boron) * 100
Let's plug in the given values:
Relative abundance of boron-10 = (10.01294 amu / 10.81 amu) * 100
Simplifying the calculation:
Relative abundance of boron-10 = 0.9255 * 100
Relative abundance of boron-10 = 92.55%
Therefore, the relative abundance of boron-10 is approximately 92.55%.
To solve this problem, you can set up an equation using the concept of weighted average.
Let's assume the relative abundance of boron-10 is represented by 'x' (as a decimal), and the relative abundance of boron-11 is represented by (1-x).
The weighted average of the isotopes' masses should equal the atomic mass of boron:
(x * mass of boron-10) + ((1 - x) * mass of boron-11) = atomic mass of boron
Plugging in the given values:
(x * 10.01294 amu) + ((1 - x) * 11.00931 amu) = 10.81 amu
Now, solve for 'x':
10.01294x + 11.00931(1 - x) = 10.81
10.01294x + 11.00931 - 11.00931x = 10.81
-0.99637x + 11.00931 = 10.81
-0.99637x = 10.81 - 11.00931
-0.99637x = -0.19931
x = (-0.19931) / (-0.99637)
x ≈ 0.2
Therefore, the relative abundance of boron-10 is approximately 0.2 (or 20%).